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Anonymous Tue 16 Mar 16:38:22 2021 No.12841500 View Reply Original Report

Quoted By: >>12843349 >>12845367

can someone explain this to me?
how do i get from cos^2(x) to 1/2(cos(2x)+1)?

Anonymous

Anonymous Tue 16 Mar 2021 17:00:25 No.12841590 Report

Quoted By:

cos(2x) = 2cos(x)cos(x)-1

Anonymous

Anonymous Tue 16 Mar 2021 22:59:44 No.12843312 Report

Quoted By: >>12843340

cos^2(x) = 1/2cos(x) + 1/2

Anonymous

Anonymous Tue 16 Mar 2021 23:05:00 No.12843327 Report

Quoted By: >>12843382

https://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/unit-4-techniques-of-integration/part-a-trigonometric-powers-trigonometric-substitution-and-completing-the-square/session-68-integral-of-sin-n-x-cos-m-x-odd-exponents/MIT18_01SCF10_Ses68a.pdf

Double angle formula can be derived from the product of two complex numbers in polar and Cartesian form.

Anonymous

Anonymous Tue 16 Mar 2021 23:08:27 No.12843340 Report

Quoted By:

>>12843312
cos^2(pi) = -1^2 = 1
1/2cos(pi) + 1/2 = -1/2 + 1/2 = 0

Anonymous

Anonymous Tue 16 Mar 2021 23:10:08 No.12843345 Report

Quoted By: >>12843382 >>12844263

you gotta know that:
cos(2x) = cos^2(x) - sin^2(x)
>use pythagorean identity
sin^2(x) + cos^2(x) = 1
cos(2x) = cos^2(x) - 1 + cos^2(x) = 2cos^2(x) - 1
cos^2(x) = cos(2x)/2 +1/2

Anonymous

Anonymous Tue 16 Mar 2021 23:11:17 No.12843349 Report

Quoted By:

>>12841500
Double angle formula. I have no idea how you niggers don't study this in hs. If you are in hs rn then fuck off this site.

Anonymous

Anonymous Tue 16 Mar 2021 23:14:53 No.12843358 Report

Quoted By:

De Moivre's gives that $\cos(2x) = \cos^2(x) - \sin^2(x) = 2\cos^2(x)-1$

Anonymous

Anonymous Tue 16 Mar 2021 23:21:32 No.12843382 Report

Quoted By:

>>12843327
>>12843345
thanks a lot

Anonymous

Anonymous Tue 16 Mar 2021 23:52:42 No.12843483 Report

Quoted By:

$\cos(2x) + i \sin(2x) = \exp(i 2x) = \exp(ix)^2 = (\cos(x) + i \sin(x))^2 = \cos^2(x) - \sin^2(x) + i 2 \cos(x) \sin(x)$

So,
$\cos(2x) = \cos^2(x) - \sin^2(x) = \cos^2(x) - (1 - \cos^2(x)) = 2 \cos^2(x) - 1$

Therefore,
$\cos^2(x) = \frac{\cos(2x) + 1}{2}$

Anonymous

Anonymous Wed 17 Mar 2021 03:08:24 No.12844263 Report

Quoted By:

>>12843345
lol, I forgot this shit twenty years ago, but this just brought it all back

Anonymous

Anonymous Wed 17 Mar 2021 11:32:49 No.12845367 Report

Quoted By:

>>12841500
cos^2(x)

is your phantasy not mathematical figure...

Also, why you don't have proper names for sides of triangle yet is what I still don't understand...

It's like side, and pythagorean and it's a lot faster your mathematics, you don't have to integrate or some simillar bullshit...

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